A pulley which is fixed to the ceiling of an elevator car carries a thread whose ends are attached to the masses $m_{1}$ and $m_{2}$ . The car starts going up with an acceleration $a_{0}$ . Assuming the masses of the pulley and the thread as well as the friction to be negligible, find:

Match the given columns according to the options where,

is the acceleration of the load $m_{1}$ relative to the ground.

Is the acceleration of the load $m_{1}$ relative to the elevator car.

is the force exerted by the pulley on the ceiling of the car. Given, $m_{1} > m_{2}$

(i) $a_{\frac{m_{1}}{g}}$ (u) $\frac{4 m_{1} m_{2} (a_{0} + g)}{(m_{1} + m_{2})}$

(ii) $a_{\frac{m_{1}}{e l e v a t o r}}$ (v) $\frac{2 a_{1} m_{1} - g m_{2}}{(m_{1} + m_{2})}$

(iii) $F_{p u l l e y o n c i e l i n g}$ (w) $\frac{2 a_{0} m_{2} - g (m_{1} - m_{2})}{(m_{1} + m_{2})}$

(i) - (x); (ii) - (w); (iii) - (u)

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(i) - (w); (ii) - (x); (iii) - (u)

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(i) - (u); (ii) - (w); (iii) - (v)

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(i) - (v); (ii) - (w); (iii) - (x)

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Solution

The correct option is B
(i) - (w); (ii) - (x); (iii) - (u)

Go in lift's frame

With respect to lift $m_{1}$ will go down and $m_{2}$ up and since string is inextensible so they will have same acceleration.

Also since you have gone to a non-inertia frame i.e. lift, you also need to take pseudo force into account.

Free Body Diagram of the blocks relative to elevator.

The elevator is an accelerated frame which is non-inertial, pseudo force $m_{1} a_{0}$ and $m_{2} a_{0}$ have been taken into consideration for $m_{1}$ and $m_{2}$ respectively.

$m_{2} g + m_{2} a_{0} - T = m_{1} a$ -------------------(i)

$T - m_{2} g - m_{2} a_{0} = m_{2} a$ --------------------(ii)

Solving the above equation, we get

$a = \frac{(m_{1} - m_{2}) . (a_{0} + g)}{(m_{1} + m_{2})}$ downward with respect to car

this is the acceleration of $m_{1}$ with respect to car.Acceleration of the mass $m_{1}$ with respect to the ground.

= $a_{0} - a_{1} = \frac{2 a_{0} m_{2} - g (m_{1} - m_{2})}{(m_{1} + m_{2})}$

Force exerted by the pulley on the celling of the car

= $2. T = (m_{1} a_{0} + m_{0} g - m_{1} a_{1}) \times 2 = \frac{4 m_{1} . m_{2} (a_{0} + g)}{(m_{1} + m_{2})}$

F = 2T

( $∵$ pulley massless)

$∴ F_{n e t} = 0$

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